How to prove the derivative of $a^{T} A^{-1} b$ with respect to $A$
$begingroup$
So I can find from the matrix cookbook here:
$frac{partial a^TX^{-1}b}{partial X} = -X^{-T}ab^TX^{-T}$
To prove it, I have tried expanding:
$a^TX^{-1}b = sumlimits_{i,j}^{n,n}a_i(X^{-1})_{ij}b_j$
I can also find from cookbook where:
$frac{partial (X^{-1})_{ij}}{partial X_{ij}} = -(X^{-1})_{ij}(X^{-1})_{ij}$
However, what I cannot figure out is that where did that transpose come from?
Any idea and help are appreciated!
matrices derivatives partial-derivative matrix-calculus
edited Dec 6 '18 at 6:41
AVK
2,0961517
asked Dec 6 '18 at 1:33
WeiWei
162
$endgroup$
add a comment |
$begingroup$
So I can find from the matrix cookbook here:
$frac{partial a^TX^{-1}b}{partial X} = -X^{-T}ab^TX^{-T}$
To prove it, I have tried expanding:
$a^TX^{-1}b = sumlimits_{i,j}^{n,n}a_i(X^{-1})_{ij}b_j$
I can also find from cookbook where:
$frac{partial (X^{-1})_{ij}}{partial X_{ij}} = -(X^{-1})_{ij}(X^{-1})_{ij}$
However, what I cannot figure out is that where did that transpose come from?
Any idea and help are appreciated!
matrices derivatives partial-derivative matrix-calculus
edited Dec 6 '18 at 6:41
AVK
2,0961517
asked Dec 6 '18 at 1:33
WeiWei
162
$endgroup$
1
$begingroup$
See here math.stackexchange.com/questions/3023692/…
$endgroup$
– user550103
Dec 6 '18 at 6:30
$begingroup$
Thank you very much for the referring. It is really helpful.
$endgroup$
– Wei
Dec 6 '18 at 23:19
add a comment |
$begingroup$
So I can find from the matrix cookbook here:
$frac{partial a^TX^{-1}b}{partial X} = -X^{-T}ab^TX^{-T}$
To prove it, I have tried expanding:
$a^TX^{-1}b = sumlimits_{i,j}^{n,n}a_i(X^{-1})_{ij}b_j$
I can also find from cookbook where:
$frac{partial (X^{-1})_{ij}}{partial X_{ij}} = -(X^{-1})_{ij}(X^{-1})_{ij}$
However, what I cannot figure out is that where did that transpose come from?
Any idea and help are appreciated!
matrices derivatives partial-derivative matrix-calculus
edited Dec 6 '18 at 6:41
AVK
2,0961517
asked Dec 6 '18 at 1:33
WeiWei
162
$endgroup$
So I can find from the matrix cookbook here:
$frac{partial a^TX^{-1}b}{partial X} = -X^{-T}ab^TX^{-T}$
To prove it, I have tried expanding:
$a^TX^{-1}b = sumlimits_{i,j}^{n,n}a_i(X^{-1})_{ij}b_j$
I can also find from cookbook where:
$frac{partial (X^{-1})_{ij}}{partial X_{ij}} = -(X^{-1})_{ij}(X^{-1})_{ij}$
However, what I cannot figure out is that where did that transpose come from?
Any idea and help are appreciated!
matrices derivatives partial-derivative matrix-calculus
matrices derivatives partial-derivative matrix-calculus
edited Dec 6 '18 at 6:41
AVK
2,0961517
asked Dec 6 '18 at 1:33
WeiWei
162
edited Dec 6 '18 at 6:41
AVK
2,0961517
asked Dec 6 '18 at 1:33
WeiWei
162
edited Dec 6 '18 at 6:41
AVK
2,0961517
edited Dec 6 '18 at 6:41
AVK
2,0961517
edited Dec 6 '18 at 6:41
AVK
2,0961517
2,0961517
asked Dec 6 '18 at 1:33
WeiWei
162
asked Dec 6 '18 at 1:33
WeiWei
162
asked Dec 6 '18 at 1:33
WeiWei
162
162
1
$begingroup$
See here math.stackexchange.com/questions/3023692/…
$endgroup$
– user550103
Dec 6 '18 at 6:30
$begingroup$
Thank you very much for the referring. It is really helpful.
$endgroup$
– Wei
Dec 6 '18 at 23:19
add a comment |
1
$begingroup$
See here math.stackexchange.com/questions/3023692/…
$endgroup$
– user550103
Dec 6 '18 at 6:30
$begingroup$
Thank you very much for the referring. It is really helpful.
$endgroup$
– Wei
Dec 6 '18 at 23:19
1
1
$begingroup$
See here math.stackexchange.com/questions/3023692/…
$endgroup$
– user550103
Dec 6 '18 at 6:30
$begingroup$
See here math.stackexchange.com/questions/3023692/…
$endgroup$
– user550103
Dec 6 '18 at 6:30
$begingroup$
Thank you very much for the referring. It is really helpful.
$endgroup$
– Wei
Dec 6 '18 at 23:19
$begingroup$
Thank you very much for the referring. It is really helpful.
$endgroup$
– Wei
Dec 6 '18 at 23:19
add a comment |
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$begingroup$
See here math.stackexchange.com/questions/3023692/…
$endgroup$
– user550103
Dec 6 '18 at 6:30
$begingroup$
Thank you very much for the referring. It is really helpful.
$endgroup$
– Wei
Dec 6 '18 at 23:19